1 + 1 + 1 + 1 + 1 = ? help me
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\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\left(\dfrac{100}{101}\right)=\dfrac{50}{101}\)
\(A=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}\)
`1/2+(-1/57)+1/3+1/6`
`=(1/2+1/3+1/6)-1/57`
`=(3/6+2/6+1/6)-1/57`
`=6/6-1/57`
`=1-1/57`
`=57/57-1/57`
`=56/57`
\(S=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{2021+2022}\)
\(S=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(S=\dfrac{1}{2}-\dfrac{1}{2022}\)
\(S=\dfrac{1011}{2022}-\dfrac{1}{2022}\)
\(S=\dfrac{505}{1011}\)
Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}$
$2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{18.19.20}$
$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+....+\frac{20-18}{18.19.20}$
$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{18.19}-\frac{1}{19.20}$
$=\frac{1}{1.2}-\frac{1}{19.20}=\frac{189}{380}$
$\Rightarrow A=\frac{189}{760}$
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{2023\times2024}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\\ =1-\dfrac{1}{2024}=\dfrac{2023}{2024}\)
1/1*2+1/2*3+...+1/2023*2024=1-1/2+1/2-1/3+...+1/2023-1/2024
=1-1/2024=2023/2024
5 nha ban k cho mk nha
1+1+1+1+1=5